thinhnghiem

Could you explain more?

Goldbach conjecture can be proved by using the Zermelo Fraenkel axioms in set theory. We apply the inductive method to prove that this conjecture is correct with all even numbers 2n with n ∈ IN. The process includes two steps 1. Prove that the conjecture is correct with n=2. 2. Assume that the conjecture is correct with a certain value n, prove that the conjecture is also correct with (n+1). We proceed with our process in sequence as below 1. With n = 2 => 2n= 4 = 2+2. The conjecture has been ascertained as correct in this case. 2. Now we assume that the conjecture is correct with n, the next step is we have to prove that it is also correct with (n+1). Here is our solution Let’s define A as the set of the odd and prime numbers x which qualify the expression x < 2n. B is also the set of the odd and prime numbers y which qualify the expression y < 2n S(x,y) is the statement to express the equation (x+y) = 2n. By the axiom schema of Replacement, there is the existence of the set B to ensure that S(x,y) is correct with at least one element x ∈ A. Since the conjecture is correct with n, we have B ≠ Ø. Let’s define A+ as the set of the odd and prime numbers x which qualify the expression x < 2(n+1). B+ is the set of the odd and prime numbers y with y < 2(n+1) and S(x,y) is correct with at least one element x ∈ A+. By the axiom of Infinity, A+ and B+ are the next items of A and B relevantly in the infinite sets. Since both A and B are ≠ Ø, then we can utilize the axiom of infinity to declare the existing of A+ and B+. That means A+ ≠ Ø and B+ ≠ Ø. The conjecture has been finally proven as correct with (n+1). By the inductive method, it is determined as correct with every n ∈ IN. Thomas Nghiem (Ontario – Canada) Reference: https://www.math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Lian.pdf

Dear all, s weknow, the movable charger for smart phone has two ports: Input and output when we do the charging by plugging the socket with the input of the charger, light turns red to show that it is in progress When the charger is full, the light turns green What happens if the input and output of the charger are connect by themselves The light turns red as if this device is being charged with power. In fact, electrical power is being discharged We never see the light turning green. Instead, when the discharging ends, light is turned off. How do we explain this phenomenon?

Dear all, I have an amusing experiment with my smart phone charger. That is, after charging it fully, I connect its input and output toghether, and get surprise result. Could you see in my youtube clip and give your explanation here url deleted Thanks Thinh Nghiem

I have an experiment to transfer solar energy to mechanical energy. First of all is a cone inspired by an axis. On the cone are two tiny cups of water. The whole system is put close to the glass in the room to get sunlight and avoid affection of wind Then, there will be interaction between sunlight and water molecules, which makes the cone rotate It is better to implement this idea in tropical countries, where there is plenty of sunlight and temperature is so high, which can heat the water cups See my clip for more detail link removed

Dear all I have used a bulb which is filled in with water, it works like a convex lens to concentrate solar energy. The performance of the water bulb gets highest level when I incline the bulb like in the clip. Solar light beam focus on a bigger dot, in compared with the dot when let the bulb standing vertically. See my clip at https://www.youtube.com/watch?v=f0Nvc6Q_Ulw

Dears, I have interesting model here I need your opinions to improve it so that it will never stop, and run permanently Thanks

Hi Studiot, I see your idea. My comparison goes with the input and output energy of the system. I know the thrush between magnets, but I consider it as INTERNAL interaction between two components of the system, not the external energy outside, so I do not involve it here. I have plan to improve my model so that it can rotate longer, for example 1 hour. Do you think it is economical. It can be used in a rotor of a generator. We do not need fuel, just our manual force, and the rotor will spin for a long time to generate current. Is it OK?

Hi all, Initial energy came from he force of my hands to turn the cylinders so that their magnets face each other. If you look into the clip, you can easily see that the cylinders rotate many rounds before stopping, which cause kinetic energy much greater than work made by me

First of all, I would like to say that this model is NOT a perpetual motion. It will stop eventually. What I want to say here is that the output useful work seems to be greater than input energy My system consists of 2 elements. Each element is a cylinder put on an axle. There are two permanent magnets stuck on each cylinder, with their north poles are faced outside. In the youtube clip that I will show you below, you can see the magnets of the first element are painted in blue, while those of the second one are crossed with X. I turn the cylinders slightly so that the north poles of the magnets are faced each other. Then, the thrust between magnets make the cylinders rotate. A single cylinder on an axle itself is not the system. It is an element of the system, which consists of 2 at all. Therefore, the thrust between magnets is not external force which affect system. It is comprehended as internal force between 2 elements of the system. The thrust from the first cylinder makes the second rotate, and the thrust from the second, in its turn, make the first rotate. Each component act as the cause to make the other rotate, and it acquires the affect from the other to rotate. While the input work of the system originates from a small force making cylinders moving short arc, and make magnets facing each other, the output dynamic energy is much higher. You can see in the clip that both cylinders rotate many circles, which create output useful work much greater than the work to make the magnets facing each other. Here is the link of my clip http://www.youtube.com/watch?v=GQUPaHE0vCg&feature=youtu.be Your comments are welcomed to determine that if energy conservation is violated in this case or not Thanks Thinh Nghiem from Vietnam

I have cut into the cables and splice the red to the green and other green to the other red. Finally I can do this with the help of a technician a) I loop the net output to input, and connect phone output to a telephone machine. No signal at all b) I loop the phone output to input, and connect net output to a modem, this time I can access internet normally Is it working correctly?

I attached my chart already

Sorry that this forum dies not allow upload iage from my machine. Thank you for recommendation

I have been considering what happens if I make the feedback from out put to input of a ADSL splitter, as well as change input to output and vice versa Below are my steps: Take the first splitter, plug the phone line to the Net output (NOT the input) of it Plug the input of the first splitter to the input of the second splitter Plug the Phone output of the second splitter to the phone output of the first one Plug the Net output of the second splitter to Modem, and connect to laptop The diagram can be found in my attached screenshot Here I made some abnormal things: Telephone signal transmits into the output, not the input of the splitter The output of second splitter is connected back to the first one Surprisingly, my laptop can access internet normally. Nothing wrong happens Why my extraordinary actions does not affect the operation of the splitters?

Thank you, I never do that